Thread: Help with volume: solid of revolution

The problem asks us to find the volume of the area bounded by y=-x^2+6x-8 and y=0 rotated around the x-axis by using the shells method.

I did this and got an answer of pi, to confirm the answer I used the disks method and got an answer of 16pi/15... please help by showing me how you do both methods.

Originally Posted by acfixerdude

The problem asks us to find the volume of the area bounded by y=-x^2+6x-8 and y=0 rotated around the x-axis by using the shells method.

I did this and got an answer of pi, to confirm the answer I used the disks method and got an answer of 16pi/15... please help by showing me how you do both methods.

You show me your work, and I'll show you mine.

Originally Posted by cribby

You show me your work, and I'll show you mine.

the graph of the original equation rotated around the x axis kinda looks like a football... if I take shells to calculate volume, then I have shells centered around x=3 and from -y to y in height. So the height is 2y and the width is going to be 2pi*r where r=3-x if we're differentiating from 2 to 3. substitute for y and we get the integral from 2 to 3 of 2(-x^2+6x-8)*2pi(3-x)dx = pi

using disks, the area of one disk is going to be pi*r^2 where r is going to be equal to y. A=pi(-x^2+6x-8)^2.... I set up the integral from 2 to 4 of pi(-x^2+6x-8)^2 dx = 16pi/15...

What am I doing wrong?

Originally Posted by acfixerdude

the graph of the original equation rotated around the x axis kinda looks like a football... if I take shells to calculate volume, then I have shells centered around x=3 and from -y to y in height. So the height is 2y and the width is going to be 2pi*r where r=3-x if we're differentiating from 2 to 3. substitute for y and we get the integral from 2 to 3 of 2(-x^2+6x-8)*2pi(3-x)dx = pi

using disks, the area of one disk is going to be pi*r^2 where r is going to be equal to y. A=pi(-x^2+6x-8)^2.... I set up the integral from 2 to 4 of pi(-x^2+6x-8)^2 dx = 16pi/15...

What am I doing wrong?

I think any "wrongness" is in the algebra of your calculation. I did this by disks and had the same set-up you describe for this method, but got

:

Originally Posted by cribby

I think any "wrongness" is in the algebra of your calculation. I did this by disks and had the same set-up you describe for this method, but got

:

I don't understand how you set yours up, thanks for trying to help out... I spoke to my professor and he says that I was mistaken in trying to differentiate with respect to x and that to do it via shells I'd have to differentiate with respect to y... I can visualize it now and know why the shells don't work with x.

Thanks